CENTER OF MASS AND MOMENT OF INERTIA 147 PROBLEMS. Find the center of mass of a uniform wire bent into the following curves: (1) An arc of a circle subtending an angle 2 0 at the center. (2) y = a sin x, between x = 0 and x = IT. (3) 7/2 = 4 ax, between x = 0 and x = 2 a. (4) The cycloid x = a (6 — sin 0), y = a (1 — cos 0), between two successive cusps. (5) Half of the cardioid r = a (1 + cos 6). 118. Center.of Mass of a Body of Any Shape and Distribution of Mass. — The illustrative examples of the last few pages are worked out by special methods in order to bring out the fact that in a great number of problems the ease with which the center of mass may be determined depends upon the choice of the element of mass. The following general expressions for an element of mass may be used whatever the . shape of the body or the distribution of its mass: (a) When the bounding surfaces of the body are given in the Cartesian coordinates the mass of an infinitesimal cube is taken as the element of mass: dm = T • dx dy dz. (b) When the bounding surfaces of the body are given in spherical coordinates the element of mass is chosen as shown in Fig. 80. In this case the following is the expression for the element of mass : dm = r • p dd • dp • p d(j> sin 0 = rp2 sin 8d6d<t> dp. (c) When the density, r, varies from point to point it is expressed in terms of the coordinates and substituted in the expression for dm. FIG. 80.